Existence of $1/i$-dense subsets Let $(X, d)$ be a compact metric space and $m$ be a Borel measure on $X$. Assume that $\lbrace A_i\rbrace$ is a nested sequence of subsets:
$\dots\subset A_i\subset\dots\subset A_2\subset A_1$ and $A_i$'s shrink to a point $x$.
Let $A$ be a closed set of positive measure and $x_i\subset\mathbb{R}$ a sequence such that $x_i\to 0$ and also we have $m(A_i\setminus A)\leq x_i$ for all $i\in \mathbb{N}$.
I want to prove that :

passing to a subsequence if necessary we can assume that $A_i\cap A$ contains a $1/i$-dense subset $E_i$ of $A_i$.

 A: Consider the following counterexample.
Let $X=[0,1]^2$ with the standard topology and Lebesgue measure. For each $n$, let $A_n=[0,1/n]\times\{0\}$. Clearly, each $A_n$ has measure zero, so $m(A_i\backslash A)\leq x_i$ for all $i\in\mathbb{N}$, no matter which nonnegative sequence $(x_i)$ and which set $A$ one chooses. So let $A=[0,1]\times [1/2,1]$. Then the distance between $A$ and $A_n$ is always $1/2$.
One can also get a counterexample in which all $A_n$ have positive measure. Just let $$A_n=\{(x,y)\in [0,1]^2: y\leq x/4, x\leq 1/n\}.$$
and keep the rest as before.

Addendum: The assumption that $x=\lim_{n\to\infty}x_n\in A$ is not sufficient either. Take again Let $X=[0,1]^2$ with the standard topology and Lebesgue measure. Let $A=[0,1/4]\times[0,1]$, $x=(0,0)$ and $A_n=\Big((1/2,1/2+1/n)\times [0,1]\Big)\cup\{x\}$. 
It works if we add the assumption that all $A_n$ are closed. Let $x$ again be the limit of $(x_n)$. Take $\epsilon>0$. Let $B_\epsilon(x)$ be the open $\epsilon$-ball around $x$. Since $(A_n)$ decreases to $\{x\}$, we have $\bigcap_n A_n\cap X\backslash B_\epsilon(x)=\emptyset$. Since this is the intersection of compact sets, by the finite intersection principle, we have some $n$ such that $A_n\cap X\backslash B_\epsilon(x)=\emptyset$. But this means $A_n\subseteq B_\epsilon(x)$ and any $\epsilon$-ball around an element of $A_n$ will contain the point $x\in A\cap A_n$. By maybe passing to  a subsequence of $(A_n)$, we can ensure that $A_n\subseteq B_{1/n}(x)$.
